+2
Planned

any plan to provide oembed api interface

Esteban Feldman 13 years ago updated by Alan Schaaf (Founder) 12 years ago 7

It will be really interesting that you also provide oEmbed api interface for your images.


http://oembed.com/#section7


Thanks!

Answer

Answer
Planned

That's a really good idea. I've just added it to my todo list. A blog post will most definitely be made when it's released, but it should be sometime next month.

Answer
Planned

That's a really good idea. I've just added it to my todo list. A blog post will most definitely be made when it's released, but it should be sometime next month.

Alan, Great! Meanwhile, since I'm working on a FOSS that is a oEmbed wrapper I will call your  http://api.imgur.com/2/image/:HASH, can I do that anonymously? Cause this will be a python lib that will provide the way to get the embedded codes for any link to imgur and other services. 

Sure, you can do that anonymously. No need to be signed in unless you want to use /2/account calls.

I meant to say if an app needs to be registered to use that. I see that can be called without that, but just to comply with your service. thx 

 Nope, it does not. API keys are only used on upload.

Great, thanks... pls let me know when you support oEmbed, to update the lib.


Hi there... another question here.


I'm working on a client side anonymous api. https://github.com/eka/thornbed/

At first it was working without hitting imgur but then I got some cases I can't resolve so when trying your ombed api I see that there is a rate limit of 500.

Is that 500 a day? an hour?

Is there any pattern you can share with me for discovering Picture ID? I have problems when I get a thumb link ex:


pic page:

http://imgur.com/0TczaPb

url:

http://i.imgur.com/0TczaPb.jpg

thumb:

http://imgur.com/0TczaPbb.jpg


So the problem there is, if my API gets any other link that's no page, I have no way to know what is the proper ID without hitting Imgur API, and that's very rate limited.

Thanks

Esteban (Eka)


The word is "ambiguous" :)